Question

A cylindrical container with a movable piston initially holds 1.5 mole of a gas at a pressure of 4 atm and a volume of 2.5 L. If the piston is moved to make volume 5L, while simultaneously withdrawing 0.75 moles of gas, what is the final pressure in atm ?

• A. 1
• B. 3
• C. 5
• D. 4

Answer 1

The correct option is A 1

Solution:- (A) $1$

From ideal gas equation,

$PV=nRT$

For different no. of moles at same temperature,

$\frac{P_1{V}_1}{n_1}=\frac{P_2{V}_2}{n_2}$

Given:-

$P_1=4atm$

$V_1=2.5L$

$n_1=1.5\text{mol}$

$P_2=P\left(\text{say}\right)=?$

$V_2=5L$

$n_2=0.75\text{mol}$

Substituting all these values in ${eq}^n\left(1\right)$, we have

$\frac{4\times 2.5}{1.5}=\frac{P\times 5}{0.75}$

$\Rightarrow P=\frac{20\times 0.75}{3\times 5}$

$\Rightarrow P=1atm$

Hence the final pressure is $1atm$.